今天连续做了三道题, 感觉越来越有感觉, 第三题直接行云流水, 10 min AC

• 203.移除链表元素 单链表巧用dummy_head删除
  • • 一开始犯的一个逻辑错误: val的节点可能连续出现
    • 解题思路
    • 代码
• 707. 设计链表 找目标位置的前一个node
  • • 解题思路
    • 代码
• 206. 反转链表
  • • 解题思路
    • 代码

203.移除链表元素 单链表巧用dummy_head删除

一开始犯的一个逻辑错误: val的节点可能连续出现

    while cur and cur.next != None:
        if cur.next.val == val:
            cur.next = cur.next.next
        cur = cur.next # 这一行错误, 应当加上else

解题思路

• 使用dummy_head:

  • 可以把目光仅聚焦在类似递推的关系上

  • 但是需要注意: 若返回head, 必须返回dummy_head.next

• 不使用dummy_head:

  • 需要用循环排除头结点是否满足删除条件

  • 同时需要注意当前节点可能是None, 循环条件是

        while cur and cur.next!=None:
    

代码

使用dummy_head

# 使用dummy_head
class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        dummy_head = ListNode(next=head)

        cur = dummy_head

        while cur.next != None:
            if cur.next.val == val:
                cur.next = cur.next.next
            else:
                cur = cur.next
        
        return dummy_head.next

不使用dummy_head

class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        cur = head

        # 如果头几个node就是val
        while cur and cur.val == val:
            cur = cur.next
        head = cur

        # 查询与删除
        while cur and cur.next != None:
            if cur.next.val == val:
                cur.next = cur.next.next
            else:
                cur = cur.next

        return head

707. 设计链表 找目标位置的前一个node

解题思路

注意对于index的delete和add, 都是找到其目标位置index前一个node

代码

class Node:
    def __init__(self, val: int, next = None):
        self.val = val
        self.next = None

class MyLinkedList:
# 一个带有dummy_head的单链表
    def __init__(self):
        self.head = Node(0)
        self.count = 0

    def get(self, index: int) -> int:
        if index >= self.count:
            return -1
        # 根据题意: index=0表示第0个
        cur = self.head.next
        for i in range(index):
            cur = cur.next
        return cur.val

    def addAtHead(self, val: int) -> None:
        obj = Node(val)
        obj.next = self.head.next
        self.head.next = obj
        self.count += 1

    def addAtTail(self, val: int) -> None:
        # 先找到末尾
        cur = self.head
        while cur.next != None:
            cur = cur.next
        # 再加入
        obj = Node(val)
        cur.next = obj
        self.count += 1

    def addAtIndex(self, index: int, val: int) -> None:
        if index > self.count:
            return -1
        # 找到index前一个
        cur = self.head
        for i in range(index):
            cur = cur.next
        # 加入
        obj = Node(val)
        obj.next = cur.next
        cur.next = obj
        self.count += 1

    def deleteAtIndex(self, index: int) -> None:
        if index >= self.count:
            return
        # 找到要删的前一个
        cur = self.head
        for i in range(index):
            cur = cur.next
        # 删除
        cur.next = cur.next.next
        self.count -= 1
        


# Your MyLinkedList object will be instantiated and called as such:
# obj = MyLinkedList()
# param_1 = obj.get(index)
# obj.addAtHead(val)
# obj.addAtTail(val)
# obj.addAtIndex(index,val)
# obj.deleteAtIndex(index)

206. 反转链表

Reverse考虑0/1的Corner Case, 存储3个指针

解题思路

Reverse

• 考虑0/1的Corner Case
• 存储3个指针: 一前, 一后, 以及后面节点的next值

代码

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        i = head

        # corner case: 数量为0或者1, 无需reverse
        if i and i.next:
            j = i.next
        else:
            return head
        
        # 原来的head, 现在的tail, next设为None
        i.next = None

        # 遍历: 记得保存j.next
        while j != None:
            t = j.next
            j.next = i
            i = j
            j = t
        
        # 最后i为head
        head = i

        return head